In the solution to this example we will use both notations for the derivative just to get you familiar with the different notations.

We are going to need the rate of change of the volume to answer these questions. This means that we will need the derivative of this function since that will give us a formula for the rate of change at any time \(t\). Now, notice that the function giving the volume of water in the tank is the same function that we saw in Example 1 in the last section except the letters have changed. The change in letters between the function in this example versus the function in the example from the last section won’t affect the work and so we can just use the answer from that example with an appropriate change in letters.

The derivative is.

Recall from our work in the first limits section that we determined that if the rate of change was positive then the quantity was increasing and if the rate of change was negative then the quantity was decreasing.

We can now work the problem.

In this case all that we need is the rate of change of the volume at \(t = 1\)or,

So, at \(t = 1\) the rate of change is negative and so the volume must be decreasing at this time.

Again, we will need the rate of change at \(t = 5\).

In this case the rate of change is positive and so the volume must be increasing at \(t = 5\).

To answer this question all that we look at is the size of the rate of change and we don’t worry about the sign of the rate of change. All that we need to know here is that the larger the number the faster the rate of change. So, in this case the volume is changing faster at \(t = 1\) than at \(t = 5\).

The volume will not be changing if it has a rate of change of zero. In order to have a rate of change of zero this means that the derivative must be zero. So, to answer this question we will then need to solve

This is easy enough to do.

So at \(t = 4\) the volume isn’t changing. Note that all this is saying is that for a brief instant the volume isn’t changing. It doesn’t say that at this point the volume will quit changing permanently.

If we go back to our answers from parts (a) and (b) we can get an idea about what is going on. At \(t = 1\) the volume is decreasing and at \(t = 5\) the volume is increasing. So, at some point in time the volume needs to switch from decreasing to increasing. That time is \(t = 4\).

This is the time in which the volume goes from decreasing to increasing and so for the briefest instant in time the volume will quit changing as it changes from decreasing to increasing.

We first need the derivative of the function and we found that in Example 3 in the last section. The derivative is,

Now all that we need is the function value and derivative (for the slope) at \(z = 3\).

The tangent line is then,

Once again, we need the derivative and we found that in Example 2 in the last section. The derivative is,

To determine if the object is moving to the right (velocity is positive) or left (velocity is negative) we need the derivative at \(t = 10\).

So, the velocity at \(t = 10\) is positive and so the object is moving to the right at \(t = 10\).

The object will stop moving if the velocity is ever zero. However, note that the only way a rational expression will ever be zero is if the numerator is zero. Since the numerator of the derivative (and hence the speed) is a constant it can’t be zero.

Therefore, the object will never stop moving.

In fact, we can say a little more here. The object will always be moving to the right since the velocity is always positive.

At first glance this seems to an all but impossible task. However, if you have some basic knowledge of the interpretations of the derivative you can get a sketch of the derivative. It will not be a perfect sketch for the most part, but you should be able to get most of the basic features of the derivative in the sketch.

Let’s start off with the following sketch of the function with a couple of additions.

Notice that at \(x = - 3\), \(x = - 1\), \(x = 2\) and \(x = 4\) the tangent line to the function is horizontal. This means that the slope of the tangent line must be zero. Now, we know that the slope of the tangent line at a particular point is also the value of the derivative of the function at that point. Therefore, we now know that,

\[f'\left( { - 3} \right) = 0\hspace{0.5in}f'\left( { - 1} \right) = 0\hspace{0.5in}f'\left( 2 \right) = 0\hspace{0.5in}f'\left( 4 \right) = 0\]

This is a good starting point for us. It gives us a few points on the graph of the derivative. It also breaks the domain of the function up into regions where the function is increasing and decreasing. We know, from our discussions above, that if the function is increasing at a point then the derivative must be positive at that point. Likewise, we know that if the function is decreasing at a point then the derivative must be negative at that point.

We can now give the following information about the derivative.

\[\begin{align*} x & < - 3 & \hspace{0.5in}f'\left( x \right) & < 0\\ - 3 < x & < - 1 & \hspace{0.5in}f'\left( x \right) & > 0\\ - 1 < x & < 2 & \hspace{0.5in}f'\left( x \right) & < 0\\ 2 < x & < 4 & \hspace{0.5in}f'\left( x \right) & < 0\\ x & > 4 & \hspace{0.5in}f'\left( x \right) & > 0\end{align*}\]

Remember that we are giving the signs of the derivatives here and these are solely a function of whether the function is increasing or decreasing. The sign of the function itself is completely immaterial here and will not in any way effect the sign of the derivative.

This may still seem like we don’t have enough information to get a sketch, but we can get a little bit more information about the derivative from the graph of the function. In the range \(x < - 3\) we know that the derivative must be negative, however we can also see that the derivative needs to be increasing in this range. It is negative here until we reach \(x = - 3\) and at this point the derivative must be zero. The only way for the derivative to be negative to the left of \(x = - 3\) and zero at \(x = - 3\) is for the derivative to increase as we increase \(x\) towards \(x = - 3\).

Now, in the range \( - 3 < x < - 1\) we know that the derivative must be zero at the endpoints and positive in between the two endpoints. Directly to the right of \(x = - 3\) the derivative must also be increasing (because it starts at zero and then goes positive – therefore it must be increasing). So, the derivative in this range must start out increasing and must eventually get back to zero at \(x = - 1\). So, at some point in this interval the derivative must start decreasing before it reaches \(x = - 1\). Now, we have to be careful here because this is just general behavior here at the two endpoints. We won’t know where the derivative goes from increasing to decreasing and it may well change between increasing and decreasing several times before we reach \(x = - 1\). All we can really say is that immediately to the right of \(x = - 3\) the derivative will be increasing and immediately to the left of \(x = - 1\) the derivative will be decreasing.

Next, for the ranges \( - 1 < x < 2\) and \(2 < x < 4\) we know the derivative will be zero at the endpoints and negative in between. Also, following the type of reasoning given above we can see in each of these ranges that the derivative will be decreasing just to the right of the left-hand endpoint and increasing just to the left of the right hand endpoint.

Finally, in the last region \(x > 4\) we know that the derivative is zero at \(x = 4\) and positive to the right of \(x = 4\). Once again, following the reasoning above, the derivative must also be increasing in this range.

Putting all of this material together (and always taking the simplest choices for increasing and/or decreasing information) gives us the following sketch for the derivative.

Note that this was done with the actual derivative and so is in fact accurate. Any sketch you do will probably not look quite the same. The “humps” in each of the regions may be at different places and/or different heights for example. Also, note that we left off the vertical scale because given the information that we’ve got at this point there was no real way to know this information.

That doesn’t mean however that we can’t get some ideas of specific points on the derivative other than where we know the derivative to be zero. To see this let’s check out the following graph of the function (not the derivative, but the function).

At \(x = - 2\) and \(x = 3\) we’ve sketched in a couple of tangent lines. We can use the basic rise/run slope concept to estimate the value of the derivative at these points.

Let’s start at \(x = 3\). We’ve got two points on the line here. We can see that each seem to be about one-quarter of the way off the grid line. So, taking that into account and the fact that we go through one complete grid we can see that the slope of the tangent line, and hence the derivative, is approximately -1.5.

At \(x = - 2\) it looks like (with some heavy estimation) that the second point is about 6.5 grids above the first point and so the slope of the tangent line here, and hence the derivative, is approximately 6.5.

Here is the sketch of the derivative with the vertical scale included and from this we can see that in fact our estimates are pretty close to reality.

Note that this idea of estimating values of derivatives can be a tricky process and does require a fair amount of (possible bad) approximations so while it can be used, you need to be careful with it.